sp³ Hybridization

We might expect the number of bonds formed by an atom to be equal to the number of unpaired electrons in its valence shell. Below is the valence shell configuration for carbon.

There are two unpaired electrons, which would indicate carbon only forms 2 bonds as in CH₂. We know that carbon will form 4 bonds. How can carbon form 4 bonds if it only has 2 unpaired electrons in its valence shell? In fact, CH₂ is very reactive and cannot be isolated. If a 2s electron is promoted to an empty 2p orbital, then the four unpaired electrons are available to form four bonds.

The four orbitals become mixed, or hybridized to form bonds that are sp³ hybridized. An sp³ hybrid oribital has two lobes with one larger than the other and they both have different phases. The larger lobes are elongated and oriented to the four corners of a tetrahedron with bond angles of 109.5^o.

In the figure below are the four sp³ orbitals. Only the larger lobes are shown.

All atoms in a molecule that have a tetrahedral electron pair geometry are sp³ hybridized. The carbon in methane is sp³ hybridized as well as the nitrogen in ammonia as shown in the figure below. Note, the lone pair of electrons on nitrogen reside in an sp³ hybridized orbital.

All four of the sp³ orbitals are identical in length, shape, and energy. Each sp³ orbital has 75% p character and 25% s character.

sp² Hybridization

If an s orbital and two p orbitals are mixed, we have sp² hybridization. The hybrid orbital has 33% s character and 67% p character. The boron atom in boron trifluoride, BF₃, is sp² hybridized. The valence shell electron configuration for boron is shown below.

Boron only has one unpaired electron in a p orbital. If an electron is promoted from the 2s orbital to a 2p orbital, there are three unpaired electrons. This will allow boron to form three bonds with sp² hybrid orbitals.

In BF₃, the electron pair geometry of boron is trigonal planar. Boron is sp² hybridized. A 2p unhybridized orbital from fluorine overlaps with an sp² hybridized orbital of boron. An atom or atoms in a molecule with an electron pair geometry of trigonal planar are sp² hybridized.

NO₂^– has a trigonal planar electron pair geometry and the nitrogen is sp² hybridized. Multiple bonds can also be sp² hybridized. For example, in CH₂O, there is a carbon-oxygen double bond. The electron pair geometry is trigonal planar.

A double bond has one sigma bond and one pi (π) bond. A π bond is formed by side to side overlap of 2 p orbitals. It is weaker than a sigma bond, because the overlap is not as strong as end-to-end overlap of p orbitals.

A π bond is weaker than a sigma bond. Generally, in chemical reactions, a π bond will break first since it takes less energy to break than it does to break a sigma bond.

sp Hybridization

Beryllium has no unpaired electrons in its ground state electron configuration.

If an electron is promoted from a 2s orbital to a 2p orbital there will be two unpaired electrons and two bonds can be formed.

An s and one p orbital have been mixed, and the result is an sp hybrid orbital. The remaining two p oribitals are empty. The number of hybridized orbitals is equal the number of orbitals that are mixed to form a hybrid orbital. In this case two sp hybrid orbitals are formed. Any atom in a molecule that has a linear geometry is sp hybridized. For example, BeCl₂ forms two bonds due to the overlap of an sp hybridized orbital on Be and a 3p unhybridized orbital of Cl.

The geometry about the Be is linear as predicted by VSEPR. The two empty p orbitals remain unhybridized. These unhybridized orbitals may be used in a multiple bond. They are useful in describing triple bonds. An sp hybrid oribital has 50% s character and 50% p character.

The hybridization of carbon in HCN is sp. The molecule is linear and there is a triple bond. A triple bond is made up of one sigma bond and two pi bonds.

Hybridization and d Orbitals

A 3s electron can be promoted to a 3d orbital which gives rise to a set of five sp³d hybrid orbitals. This is typical for Group 5A elements. One example is phosphorous. Below is the valence shell orbital box diagram of phosphorous.

If one electron from the 3s orbital is promoted to a 3d orbital, there are five sp³d hybrid orbitals.

This corresponds to a trigonal bipyramidal electron pair geometry. An example is PCl₅. The overlap is between the sp³d hybrid orbitals of phosphorous and a 3p orbital of chlorine.

Finally, for an octahedral electron pair geometry, sp³d² hybrid orbitals are formed. This mainly occurs for Group 6A elements below the second period. One example is sulfur. The valence orbital diagram for sulfur is shown below.

If an electron is promoted from a 3s orbital and one from a 3p orbital, there are 6 hybrid orbitals, sp³d².

The hybridization model works well for atoms that have an octet of electrons, but not as well for those with more than an octet of electrons. These more advanced models are taught in a higher level course, and are above the level of general chemistry.

In order to determine the hybridization of an atom, first draw the Lewis structure. Then use VSEPR to determine the electron pair geometry. You can then assign sp, sp², or sp³ hybridization.
 

Worksheet: Valence Bond Theory and Hybridization

Exercises

Exercise 1. Which is a stronger bond, a σ or π bond?

Check Answer/Solution to Exercise 1

Exercise 2. What hybridization is expected for the underlined atoms in the following?

Check Answer/Solution to Exercise 2

Exercise 3. What is the hybridization of Br in BrO₂^–?

Check Answer/Solution to Exercise 3

Exercise 4. What is the hybridization of both N atoms in H₂N₂?

Check Answer/Solution to Exercise 4

Exercise 5. Describe the overlap of orbitals for H₃C-CN as well as the hybridization of the two carbon atoms.

Check Answer/Solution to Exercise 5

Exercise 6. Describe the overlap of orbitals in HCCCl. Use hybridization schemes.

Check Answer/Solution to Exercise 6

Exercise 7. The following shows how the atoms are connected in the amino acid leucine.

Draw a Lewis structure and then determine the hybridization of all non-terminal atoms.

Check Answer/Solution to Exercise 7

 
Back to Bonding Models for Covalent Compounds
Back to General Chemistry 1 Study Guides
Back to Home Page